J89's pass-for-ko: Misinterpreted in the Western Go World?

Cassandra · **#41**

kvasir wrote:

But in example 18 it appears that unless pass-ko is limited (for example as Jann suggests) it is unclear why the marked stone is dead.

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O . O X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O Y . X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

can be captured, but will be never re-established.

---------------------------

Click Here To Show Diagram Code: [go]$$W $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O 2 O X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O X 1 X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$W pass $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O X 5 X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O 4 O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$W pass $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O 8 O X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O X 7 X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

---------------------------

Variation:

Click Here To Show Diagram Code: [go]$$W $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O 2 O X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O X 1 X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$W pass-for-ko; pass-for-ko; pass $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O X 7 X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O 6 O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

Same as above.

Cassandra · **#42**

kvasir wrote:

Click Here To Show Diagram Code: [go]$$W $$ +---------------------------+ $$ | . 7 . 4 3 2 . 1 1 1 1 1 1 | $$ | 5 4 4 4 3 2 . 1 1 1 1 1 1 | $$ | . 4 3 3 3 2 2 2 2 2 2 2 2 | $$ | 4 4 3 6 6 O O O O O O O O | $$ | 3 3 3 6 O O . . . . . . . | $$ | . . 6 6 O . . . . . . . . | $$ | 6 6 6 O O . , . . , . . . | $$ | O O O O . . . . . . . . . | $$ | . . . . . . . . . . . . . | $$ | . . . , . . , . . , . . . |[/go]

I don't think this is controversial or unexpected:
:w1:

seki,

seki,

alive,

dead,

alive,

dead,

alive

Sorry, but...

++++++++++++++++++++++++++++++++

Status confirmation WITHOUT any kind of ko ban:

Click Here To Show Diagram Code: [go]$$W $$ +---------------------------+ $$ | . W . # W # . W W W W W W | $$ | W # # # W # . W W W W W W | $$ | . # W W W # # # # # # # # | $$ | # # W Z Z O O O O O O O O | $$ | W W W Z O O . . . . . . . | $$ | . . Z Z O . . . . . . . . | $$ | Z Z Z O O . , . . , . . . | $$ | O O O O . . . . . . . . . | $$ | . . . . . . . . . . . . . | $$ | . . . , . . , . . , . . . |[/go]

Click Here To Show Diagram Code: [go]$$W $$ +---------------------------+ $$ | . @ . # W # . @ @ @ @ @ @ | $$ | @ # # # W # . @ @ @ @ @ @ | $$ | . # W W W # # # # # # # # | $$ | # # W Z Z O O O O O O O O | $$ | W W W Z O O . . . . . . . | $$ | . . Z Z O . . . . . . . . | $$ | Z Z Z O O . , . . , . . . | $$ | O O O O . . . . . . . . . | $$ | . . . . . . . . . . . . . | $$ | . . . , . . , . . , . . . |[/go]

Black's groups in the corners are "alive in seki".
White's groups in the corners are "alive in seki".
White's group that surrounds the bent-four is "alive".
Black's group at the left edge is "dead".

Thus, the region at the left edge is White territory.

--------------------------------

In the case "re-establishment" of captured stones is restricted to the board points that were former occupied by these:

Click Here To Show Diagram Code: [go]$$W $$ +---------------------------+ $$ | . W . Z W Z . W W W W W W | $$ | W Z Z Z W Z . W W W W W W | $$ | . Z W W W Z Z Z Z Z Z Z Z | $$ | Z Z W Z Z O O O O O O O O | $$ | W W W Z O O . . . . . . . | $$ | . . Z Z O . . . . . . . . | $$ | Z Z Z O O . , . . , . . . | $$ | O O O O . . . . . . . . . | $$ | . . . . . . . . . . . . . | $$ | . . . , . . , . . , . . . |[/go]

All White groups are "alive".
All Black groups are "dead".

++++++++++++++++++++++++++++++++

J89's status confirmation with "pass-for-ko":

Click Here To Show Diagram Code: [go]$$W $$ +---------------------------+ $$ | . W . Z W Z . W W W W W W | $$ | W Z Z Z W Z . W W W W W W | $$ | . Z W W W Z Z Z Z Z Z Z Z | $$ | Z Z W Z Z O O O O O O O O | $$ | W W W Z O O . . . . . . . | $$ | . . Z Z O . . . . . . . . | $$ | Z Z Z O O . , . . , . . . | $$ | O O O O . . . . . . . . . | $$ | . . . . . . . . . . . . . | $$ | . . . , . . , . . , . . . |[/go]

All White groups are "alive".
All Black groups are "dead".

kvasir · **#43**

Cassandra wrote:

J89's status confirmation with "pass-for-ko":

Click Here To Show Diagram Code: [go]$$W $$ +---------------------------+ $$ | . W . Z W Z . W W W W W W | $$ | W Z Z Z W Z . W W W W W W | $$ | . Z W W W Z Z Z Z Z Z Z Z | $$ | Z Z W Z Z O O O O O O O O | $$ | W W W Z O O . . . . . . . | $$ | . . Z Z O . . . . . . . . | $$ | Z Z Z O O . , . . , . . . | $$ | O O O O . . . . . . . . . | $$ | . . . . . . . . . . . . . | $$ | . . . , . . , . . , . . . |[/go]

All White groups are "alive".
All Black groups are "dead".

I wasn't sure what your conclusion about the group of white stones in the top-right was. It is seki but you seem to say the opposite that black is dead, but that can only be the case if black had to remove a liberty.

Click Here To Show Diagram Code: [go]$$B black never plays $$ +---------------------------+ $$ | . W . Z W Z . W W W W W W | $$ | W Z Z Z W Z 1 W W W W W W | $$ | . Z W W W Z Z Z Z Z Z Z Z | $$ | Z Z W Z Z O O O O O O O O | $$ | W W W Z O O . . . . . . . | $$ | . . Z Z O . . . . . . . . | $$ | Z Z Z O O . , . . , . . . | $$ | O O O O . . . . . . . . . | $$ | . . . . . . . . . . . . . | $$ | . . . , . . , . . , . . . |[/go]

Cassandra wrote:

kvasir wrote:

But in example 18 it appears that unless pass-ko is limited (for example as Jann suggests) it is unclear why the marked stone is dead.

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O . O X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O Y . X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

can be captured, but will be never re-established.

OK. I realized that the rules are quite clear on this stone being dead. "Life-and-Death Example 25: Double-Ko Seki" also explains that the ko stones in double-ko are dead because they can always be recaptured, so that is clear.

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O . W X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O X . X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

If this is the case then why isn't the marked stone dead? Same argument as before, pass-ko allows black to recapture the ko indefinitely.

This example seems to get to the core of the pass-ko interpretation.

Cassandra · **#44**

kvasir wrote:

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O . W X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O X . X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .| $$ | . . . . . . . . . . .|[/go]

If this is the case then why isn't the marked stone dead? Same argument as before, pass-ko allows black to recapture the ko indefinitely.

This example seems to get to the core of the pass-ko interpretation.

Click Here To Show Diagram Code: [go]$$B pass $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O 1 O X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O X 2 X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | 4 O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O X C X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O . O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Thereafter, White kills everything in the corner under J89.

:ec:

will become :wc:

again.

It's a bit tricky, I know. But there can be different sequences for different groups under consideration.

kvasir · **#45**

Cassandra wrote:

It's a bit tricky, I know. But there can be different sequences for different groups under consideration.

If there are two different determinations for the same stone, it is actually not "tricky" it is "indeterminate".

You had this diagram and I add some moves, noting that while black did pass for the ko, white didn't and therefore can't retake.

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | 4 O 6 X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O X . X X O , .| $$ | O O O 7 O X . X O . .| $$ | X X O O 5 O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Basically, if white had passed for a ko then :w4:

couldn't have been played. This line of reasoning does obviously not match the determination in the examples. Anyway, we did discus some possible reasons already.

Cassandra · **#46**

kvasir wrote:

Cassandra wrote:

It's a bit tricky, I know. But there can be different sequences for different groups under consideration.

If there are two different determinations for the same stone, it is actually not "tricky" it is "indeterminate".

You had this diagram and I add some moves, noting that while black did pass for the ko, white didn't and therefore can't retake.

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | 4 O 6 X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O X . X X O , .| $$ | O O O 7 O X . X O . .| $$ | X X O O 5 O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Basically, if white had passed for a ko then :w4:

couldn't have been played. This line of reasoning does obviously not match the determination in the examples. Anyway, we did discus some possible reasons already.

"Indeterminate" ...

A very interesting thought!

In the "uncapturable" world, where J89 are a part of, the single stones in the double-ko are "dead" -- in principle -- due to the double-ko cycle.
"Usually" this does not matter, as BOTH of the large groups of the double-ko are "alive".

But in L&D example 18, where the double-ko is utilised as an example of an irremovable ko-threat, ONE of the large double-ko groups is wanted to be "dead". Consequently, there must exist another (competing???) sequence than the double-ko cycle that has the single stone re-established at its end.
Otherwise, the large "dead"-wanted group of the double-ko would not be completely surrounded by "alive" White stones.

This might be another reason why NO explicite sequence for status confirmation is shown.
Additionally, this "blank sheet" also avoids opening Pandora's Box of "ko-capture, ko-capture, pass-for-ko, pass-for-ko" for the double-ko.

Cassandra · **#47**

kvasir wrote:

You had this diagram and I add some moves, noting that while black did pass for the ko, white didn't and therefore can't retake.

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | 4 O 6 X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O X . X X O , .| $$ | O O O 7 O X . X O . .| $$ | X X O O 5 O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Basically, if white had passed for a ko then :w4:

couldn't have been played. This line of reasoning does obviously not match the determination in the examples. Anyway, we did discus some possible reasons already.

Under J89^{(<= 2007)}

Click Here To Show Diagram Code: [go]$$B pass-for-ko $$ +---------------------–- $$ | . O . X O X O . . . .| $$ | O X X X O X O O . . .| $$ | . X O O O X X O O . .| $$ | X X O O 1 O X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O X 2 X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$B pass $$ +---------------------–- $$ | 4 O 6 X O X O . . . .| $$ | O X X X O X O O . . .| $$ | 7 X O O O X X O O . .| $$ | X X O O X M X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O T O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Black must not recapture at :et:

, as long as any :ex:

is still on the board.

Click Here To Show Diagram Code: [go]$$Wm8 pass; pass $$ +---------------------–- $$ | 3 1 6 X O X O . . . .| $$ | 5 X X X O X O O . . .| $$ | X X O O O X X O O . .| $$ | X X O O X M X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O T O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$Wm14 pass $$ +---------------------–- $$ | 1 3 X X O X O . . . .| $$ | 4 X X X O X O O . . .| $$ | X X O O O X X O O . .| $$ | X X O O X M X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O T O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$Wm14 pass $$ +---------------------–- $$ | 5 7 X X O X O . . . .| $$ | X X X X O X O O . . .| $$ | X X O O O X X O O . .| $$ | X X O O X M X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O T O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$Wm22 pass; pass; pass-for-ko $$ +---------------------–- $$ | O O . . O X O . . . .| $$ | . . . . O X O O . . .| $$ | . . O O O X X O O . .| $$ | . . O O X M X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O 1 O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$Wm22 $$ +---------------------–- $$ | O O . . O X O . . . .| $$ | . . . . O X O O . . .| $$ | . . O O O X X O O . .| $$ | . . O O X . X X O , .| $$ | O O O . O X . X O . .| $$ | X X O O O O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$Wm22 pass; pass-for-ko $$ +---------------------–- $$ | O O . . O X O . . . .| $$ | . . . . O X O O . . .| $$ | . . O O O X X O O . .| $$ | . . O O X 5 X X O , .| $$ | O O O . O X 7 X O . .| $$ | X X O O O O X X O . .| $$ | . X X O O X X O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

Click Here To Show Diagram Code: [go]$$Wm22 pass; pass-for-ko $$ +---------------------–- $$ | O O . . O . O . . . .| $$ | . . . . O . O O . . .| $$ | . . O O O . . O O . .| $$ | . . O O . O . . O , .| $$ | O O O . O . O . O . .| $$ | X X O O O O . . O . .| $$ | . X X O O . . O O . .| $$ | . . X X X O O O . . .| $$ | . . . . X X X X . . .| $$ | . . . , . . . . . , .| $$ | . . . . . . . . . . .|[/go]

jann · **#48**

Cassandra wrote:

But in L&D example 18, where the double-ko is utilised as an example of an irremovable ko-threat, ONE of the large double-ko groups is wanted to be "dead". Consequently, there must exist another (competing???) sequence than the double-ko cycle that has the single stone re-established at its end.
Otherwise, the large "dead"-wanted group of the double-ko would not be completely surrounded by "alive" White stones.

This might be another reason why NO explicite sequence for status confirmation is shown.
Additionally, this "blank sheet" also avoids opening Pandora's Box of "ko-capture, ko-capture, pass-for-ko, pass-for-ko" for the double-ko.

That box (and the closed double ko cycle) definitely doesn't exist in J89 eyes, this is clear from the text and examples. The two possible explanations are here.

Whichever of the two is correct, #18 is meant to work and show that pass-for-ko allows W to capture the corner and consequently the right, with a sequence in which it answers all B "threats", moves, ko-passes on the two side separately, maintaining the temporary double ko seki for a while (not even "enabling" B to transform it into independent life, which is the threat missing from the earlier, defective example).

This is J89's intent, and this is achieved in both explanations linked above.

jann · **#49**

Gérard TAILLE wrote:

"pass once" and "pass each time" for a ko would differ, and this position is not a real moonshine life.
In addition when I analyse this position with GT territory rule it is clear that the board is not black territory. The reason is quite simple; the game is not finished and white should not pass in normal play in this position.

I guess we can find a finished position with the same kind of result (a loop reappearing) but it will surely take some time to find it. ;-)

Actually constructing such a finished example could be valuable. Theoretically possible, in practice in finished positions - not sure. But for the moment, this "pass once per ko" interpretation may be the simplest option to use/study J89 without the double ko flaw.

I also wonder what Davies had in mind when he wrote "after passing ONCE for that particular ko capture" (which seems less emphasized in Japanese). Had he some further knowledge or consulted J89 authors? The latter may be possible even now, I suppose there are ways to request clarification from the Nihon Ki-in - just nobody did this for 30 years.

Cassandra · **#50**

Dear jann,

Just returned from walking the labrador ...

Obviously, my subconscious mind did not find your assessments soooo absurd (while my consciousness prefers clear structures and rules).
Probably it found the crucial point in James Davies' translation (we do not know whether is was also contained in the original version at the time).

But first let me digress a little for a better understanding.

What is Igo Hatsuyôron good for?

Inoue Dôsetsu Inseki (1713) wrote:

Dear students, in our former lessons, I used simple, basic examples to show you techniques (aka "principles") how you can kill opposing groups and save your own.
Here are some sample applications that you can use to find out for yourself how to correctly use these techniques to achieve the stated results.
Solutions not included!

What are J89's L&D Examples 16 to 25 good for?

J89 (version unknown) wrote:

Dear Go players, we used the simple, basic examples 1 to 15 to show you principles how you can identify the L&D-status of a given formation.
Here are some sample applications that you can use to find out for yourself how to correctly use these principles to achieve the stated results.
Solutions not included!

+++++++++++++++++++++++++++++++

James Davies' translation	Translation of CURRENT version
EXAMPLES OF CONFIRMATION OF LIFE AND DEATH	Life and Death Confirmation Examples
The results in the following examples would be reached through confirmation of life and death if the game stopped in the position shown in the diagram.	The judgement of "life and death" of the following life and death examples IS with the status of the examplary diagrams after the stop of the game.
They do not prevent these positions from being resolved through actual play before the end of the game.	They are judgement results when life and death is confirmed, and do not prevent them from being resolved through actual play before the end of the game.

______________________
^{Emphasis mine}

Additional remark: "confirmation" leads to another family of Japanese kanji than "judgement".

Gérard TAILLE · **#51**

jann wrote:

Gérard TAILLE wrote:

"pass once" and "pass each time" for a ko would differ, and this position is not a real moonshine life.
In addition when I analyse this position with GT territory rule it is clear that the board is not black territory. The reason is quite simple; the game is not finished and white should not pass in normal play in this position.

I guess we can find a finished position with the same kind of result (a loop reappearing) but it will surely take some time to find it. ;-)

Actually constructing such a finished example could be valuable. Theoretically possible, in practice in finished positions - not sure. But for the moment, this "pass once per ko" interpretation may be the simplest option to use/study J89 without the double ko flaw.

I also wonder what Davies had in mind when he wrote "after passing ONCE for that particular ko capture" (which seems less emphasized in Japanese). Had he some further knowledge or consulted J89 authors? The latter may be possible even now, I suppose there are ways to request clarification from the Nihon Ki-in - just nobody did this for 30 years.

For the time being I have still no idea to construct such unfinished position but I am also unable to prove that such position cannot exist ;-)

BTW, even if "pass once per ko" works isn't a good other solution to take instead the "ko-pass" (definied in J2003) because it looks still simplier?

Cassandra · **#52**

Gérard TAILLE wrote:

jann wrote:

Gérard TAILLE wrote:

"pass once" and "pass each time" for a ko would differ, and this position is not a real moonshine life.
In addition when I analyse this position with GT territory rule it is clear that the board is not black territory. The reason is quite simple; the game is not finished and white should not pass in normal play in this position.

I guess we can find a finished position with the same kind of result (a loop reappearing) but it will surely take some time to find it. ;-)

Actually constructing such a finished example could be valuable. Theoretically possible, in practice in finished positions - not sure. But for the moment, this "pass once per ko" interpretation may be the simplest option to use/study J89 without the double ko flaw.

I also wonder what Davies had in mind when he wrote "after passing ONCE for that particular ko capture" (which seems less emphasized in Japanese). Had he some further knowledge or consulted J89 authors? The latter may be possible even now, I suppose there are ways to request clarification from the Nihon Ki-in - just nobody did this for 30 years.

For the time being I have still no idea to construct such unfinished position but I am also unable to prove that such position cannot exist ;-)

BTW, even if "pass once per ko" works isn't a good other solution to take instead the "ko-pass" (definied in J2003) because it looks still simplier?

English "once" = Japanese "一度"; "一時"; "一応"; "一遍"; "一頃"; "往年" does NOT appear in the current legal text of J89!

jann · **#53**

The Japanese sentence has other structure than Davies' translation, it says sth like "whichever ko recapture (he) passed for respectively, THAT ko can be taken afterward anew". But neither of us has good enough Japanese, so..

jann · **#54**

Gérard TAILLE wrote:

BTW, even if "pass once per ko" works isn't a good other solution to take instead the "ko-pass" (definied in J2003) because it looks still simplier?

Again, think in logical concepts. J89 ko rule (w/o double ko issue) has theoretical meaning of try to separate kos from each other, and play them independently. A global ko pass that unlocks all kos for all players at once has no such conceptual ground just unpredictable effects.

J2003 may or may not get away with this by patching together Korean localization and Japanese-like ko passes and local play/enable, but J89 has global play and global enable so I doubt this would work well. See J89 LD 7.2 example.

kvasir · **#55**

One could also treat confirmation as an iterative process as follows:

1. Determine for every string of stones if they can be captured, and not reestablished with stones that can not be captured, and declare these stones to be alive.
2. Remove stones that are declared alive in step 1. from further consideration in step 1.
3. Remove stones (or treat as if removed) from the board that are both not alive and surrounded only by alive stones of the other color.
4. Repeat from step 1. until no more stones are removed in step 3.

Technically one may need to address infinite cycles that could be used to prevent captures, but by the handwaving method these cases should at the end of the loop only affect stones that are in double-ko-like seki (i.e. sekis that have dead stones).

This does have limited support in the examples which talk about "collapse of the seki" in various places, including in example 18, but the rules do otherwise not imply any iteration during confirmation. However, this seems to match quite well what happens when a game is finished.

...or does this have some obvious flaws?

Gérard TAILLE · **#56**

kvasir wrote:

One could also treat confirmation as an iterative process as follows:

1. Determine for every string of stones if they can be captured, and not reestablished with stones that can not be captured, and declare these stones to be alive.
2. Remove stones that are declared alive in step 1. from further consideration in step 1.
3. Remove stones (or treat as if removed) from the board that are both not alive and surrounded only by alive stones of the other color.
4. Repeat from step 1. until no more stones are removed in step 3.

Technically one may need to address infinite cycles that could be used to prevent captures, but by the handwaving method these cases should at the end of the loop only affect stones that are in double-ko-like seki (i.e. sekis that have dead stones).

This does have limited support in the examples which talk about "collapse of the seki" in various places, including in example 18, but the rules do otherwise not imply any iteration during confirmation. However, this seems to match quite well what happens when a game is finished.

...or does this have some obvious flaws?

It seems there is some wording problems especially in your first point
1. Determine for every string of stones if they can be captured, and not reestablished with stones that can not be captured, and declare these stones to be ALIVE.

Anyway maybe can you show your proposal on the very first L&D example 1

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . X X O . . . $$ | O X X O . . . $$ | X O O O . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Gérard TAILLE · **#57**

jann wrote:

Gérard TAILLE wrote:

BTW, even if "pass once per ko" works isn't a good other solution to take instead the "ko-pass" (definied in J2003) because it looks still simplier?

Again, think in logical concepts. J89 ko rule (w/o double ko issue) has theoretical meaning of try to separate kos from each other, and play them independently. A global ko pass that unlocks all kos for all players at once has no such conceptual ground just unpredictable effects.

J2003 may or may not get away with this by patching together Korean localization and Japanese-like ko passes and local play/enable, but J89 has global play and global enable so I doubt this would work well. See J89 LD 7.2 example.

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . X X O X O . . . O X O . O . | $$ | X O . O X O . . . O X X O O O | $$ | O O X X X O . . . O X . X O X | $$ | O X X O O O . . . O X X X X . | $$ | . O X O . . . . . O O O O X X | $$ | a O X O . . . . . . . . O O O | $$ | O X X O . . . . . . . . . . . | $$ | O O O O . . . . . . . . . . . | $$ | . . . . . . . . . . . . . . . | $$ | . . . . . . . . . . . . . . . |[/go]

It is black to play and we are still in normal play.
Assume that if black passes and white adds a move at "a" then black wins the game by 0.5 points.

jann wrote:

This is a longstanding question in J89 which has two possible answers:

Maybe passing for each ko is only required once for each player (see here)
Or the problem was overlooked by J89 authors, and will be fixed in the future (with some hack like requiring a normal pass of either player before passing for a ko a second time)

1)

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . X X W X O . . . O X O . O . | $$ | X O . W X O . . . O X X O O O | $$ | O O X X X O . . . O X . X O X | $$ | O X X O O O . . . O X X X X . | $$ | . O X O . . . . . O O O O X X | $$ | . O X O . . . . . . . . O O O | $$ | O X X O . . . . . . . . . . . | $$ | O O O O . . . . . . . . . . . | $$ | . . . . . . . . . . . . . . . | $$ | . . . . . . . . . . . . . . . |[/go]

If you use "pass once per ko" then black can pass immediatly and wins the game because the two white marked stones are dead in the confirmation phase

2)

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | 2 X X O X O . . . O X O . O . | $$ | X O 1 O X O . . . O X X O O O | $$ | O O X X X O . . . O X . X O X | $$ | O X X O O O . . . O X X X X . | $$ | . O X O . . . . . O O O O X X | $$ | . O X O . . . . . . . . O O O | $$ | O X X O . . . . . . . . . . . | $$ | O O O O . . . . . . . . . . . | $$ | . . . . . . . . . . . . . . . | $$ | . . . . . . . . . . . . . . . |[/go]

without using "pass once per ko", in order to win, black must continue at least one move.
After :b1:

then black passes and wins the game.

As you see using "pass once per ko" may change normal play.

Alternating systematically a "ko ban requiring an explicit pass-for-ko" and a "normal ko ban" seems to resolve the problem.

kvasir · **#58**

Gérard TAILLE wrote:

It seems there is some wording problems especially in your first point
1. Determine for every string of stones if they can be captured, and not reestablished with stones that can not be captured, and declare these stones to be ALIVE.

Anyway maybe can you show your proposal on the very first L&D example 1

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . X X O . . . $$ | O X X O . . . $$ | X O O O . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Sorry about the bad wording. My intention was to convey the general idea.

It was the intention that step 1 was the regular j89 determination of alive status. In example 1 all the stones are alive, and there is nothing different from the non-iterative process.

I notice that I should have said for step 3 that the dead stones are to be removed if they are surrounded only by dead stones of the same color or alive stones of the other color. I hope this is understandable enough if a little bit imprecise.

Another issue is the (deliberately) imprecise treatment of removed dead stones in step 3, I think a little bit more needs to be said about how to treat them in later iteration. For now I am assuming (somewhat contrary to what I said before) that they are left in-place, can be captured by the other color, but the same color can not play new stones in place of captured stones that were marked as dead.

The first example of iterative application would be example 8, so I will try to demonstrate with this example.

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | X O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | O X . X O . . $$ | . O X O X . . $$ | O O O O X . . $$ | . O X X X . . $$ | O O X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Iteration 1

Step 1 determination of alive status.

Click Here To Show Diagram Code: [go]$$B step 1 $$ | . . . . . . . $$ | O O . . . . . $$ | 1 O . . . . . $$ | . 1 O O O . . $$ | 1 1 1 1 O . . $$ | 2 1 . 1 O . . $$ | . 4 1 4 X . . $$ | 4 4 4 4 X . . $$ | . 4 X X X . . $$ | 4 4 X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Statuses:

dead - per the original rules
:w2:

dead - assuming double-ko "abuse" because black can capture this stone in a cycle.
:w4:

alive - per the original rules

This would be sufficient, if double-ko "abuse" is ignored, but here the loop continues.

In step 2 the alive white stones are removed from further consideration, marked with x-es.

Click Here To Show Diagram Code: [go]$$B step 2 $$ | . . . . . . . $$ | O O . . . . . $$ | X O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | O X . X O . . $$ | . P X P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

In step 3 there is one black stone to mark as dead.

Click Here To Show Diagram Code: [go]$$B step 3 $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | O X . X O . . $$ | . P X P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Iteration 2

Click Here To Show Diagram Code: [go]$$B step 1 $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . 1 O O O . . $$ | 1 1 1 1 O . . $$ | 2 1 . 1 O . . $$ | . P 1 P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Now that some stones fixed as dead or alive we have the following statuses for the remaining stones:

:b1:

dead

alive

In step 2 the alive white stones are removed from further consideration, marked with x-es.

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | P X . X O . . $$ | . P X P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

In step 3 the dead black stones that are surrounded by alive white stones and dead black stones are marked as dead.

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . A O O O . . $$ | A A A A O . . $$ | P A . A O . . $$ | . P A P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Iteration 3

There are no stones left relating to the example so the loop terminates with no chances from the previous diagram.

Status at end of the loop

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | 1 O . . . . . $$ | . 1 O O O . . $$ | 1 1 1 1 O . . $$ | 2 1 . 1 O . . $$ | . 2 1 2 X . . $$ | 2 2 2 2 X . . $$ | . 2 X X X . . $$ | 2 2 X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

dead

alive

I hope this explains the general concept, even if I have not come up with a precise enough formulation yet.

Gérard TAILLE · **#59**

kvasir wrote:

Gérard TAILLE wrote:

It seems there is some wording problems especially in your first point
1. Determine for every string of stones if they can be captured, and not reestablished with stones that can not be captured, and declare these stones to be ALIVE.

Anyway maybe can you show your proposal on the very first L&D example 1

Click Here To Show Diagram Code: [go]$$B $$ +---------------------–- $$ | . X X O . . . $$ | O X X O . . . $$ | X O O O . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Sorry about the bad wording. My intention was to convey the general idea.

It was the intention that step 1 was the regular j89 determination of alive status. In example 1 all the stones are alive, and there is nothing different from the non-iterative process.

I notice that I should have said for step 3 that the dead stones are to be removed if they are surrounded only by dead stones of the same color or alive stones of the other color. I hope this is understandable enough if a little bit imprecise.

Another issue is the (deliberately) imprecise treatment of removed dead stones in step 3, I think a little bit more needs to be said about how to treat them in later iteration. For now I am assuming (somewhat contrary to what I said before) that they are left in-place, can be captured by the other color, but the same color can not play new stones in place of captured stones that were marked as dead.

The first example of iterative application would be example 8, so I will try to demonstrate with this example.

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | X O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | O X . X O . . $$ | . O X O X . . $$ | O O O O X . . $$ | . O X X X . . $$ | O O X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Iteration 1

Step 1 determination of alive status.

Click Here To Show Diagram Code: [go]$$B step 1 $$ | . . . . . . . $$ | O O . . . . . $$ | 1 O . . . . . $$ | . 1 O O O . . $$ | 1 1 1 1 O . . $$ | 2 1 . 1 O . . $$ | . 4 1 4 X . . $$ | 4 4 4 4 X . . $$ | . 4 X X X . . $$ | 4 4 X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Statuses:

dead - per the original rules
:w2:

dead - assuming double-ko "abuse" because black can capture this stone in a cycle.
:w4:

alive - per the original rules

This would be sufficient, if double-ko "abuse" is ignored, but here the loop continues.

In step 2 the alive white stones are removed from further consideration, marked with x-es.

Click Here To Show Diagram Code: [go]$$B step 2 $$ | . . . . . . . $$ | O O . . . . . $$ | X O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | O X . X O . . $$ | . P X P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

In step 3 there is one black stone to mark as dead.

Click Here To Show Diagram Code: [go]$$B step 3 $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | O X . X O . . $$ | . P X P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Iteration 2

Click Here To Show Diagram Code: [go]$$B step 1 $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . 1 O O O . . $$ | 1 1 1 1 O . . $$ | 2 1 . 1 O . . $$ | . P 1 P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Now that some stones fixed as dead or alive we have the following statuses for the remaining stones:

:b1:

dead

alive

In step 2 the alive white stones are removed from further consideration, marked with x-es.

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . X O O O . . $$ | X X X X O . . $$ | P X . X O . . $$ | . P X P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

In step 3 the dead black stones that are surrounded by alive white stones and dead black stones are marked as dead.

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | A O . . . . . $$ | . A O O O . . $$ | A A A A O . . $$ | P A . A O . . $$ | . P A P X . . $$ | P P P P X . . $$ | . P X X X . . $$ | P P X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

Iteration 3

There are no stones left relating to the example so the loop terminates with no chances from the previous diagram.

Status at end of the loop

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | 1 O . . . . . $$ | . 1 O O O . . $$ | 1 1 1 1 O . . $$ | 2 1 . 1 O . . $$ | . 2 1 2 X . . $$ | 2 2 2 2 X . . $$ | . 2 X X X . . $$ | 2 2 X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

dead

alive

I hope this explains the general concept, even if I have not come up with a precise enough formulation yet.

It seems not clear how you handle this famous double-ko "abuse"
Let's take a slightly different position:

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | a O . . . . . $$ | X O . . . . . $$ | . B O O O . . $$ | B B B B O . . $$ | O B . B O . . $$ | . O X O X . . $$ | O O O O X . . $$ | . O X X X . . $$ | O O X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

In this position, due to the liberty at "a" and the double-ko "abuse" then black marked stones become uncapturable (=> alive). How do you correct this result?

Cassandra · **#60**

Gérard TAILLE wrote:

It seems not clear how you handle this famous double-ko "abuse"
Let's take a slightly different position:

Click Here To Show Diagram Code: [go]$$B $$ | . . . . . . . $$ | O O . . . . . $$ | a O . . . . . $$ | X O . . . . . $$ | . B O O O . . $$ | B B B B O . . $$ | O B . B O . . $$ | . O X O X . . $$ | O O O O X . . $$ | . O X X X . . $$ | O O X . . . . $$ | X X X . . . . $$ | . . . . . . .[/go]

In this position, due to the liberty at "a" and the double-ko "abuse" then black marked stones become uncapturable (=> alive). How do you correct this result?

Please be so very kind to discuss this further in your own thread "Japonese counting" or whereever.
But not here. You are derailing this thread!

There is NO "double-ko abuse" within the current J89!!!

Thank you.

J89's pass-for-ko: Misinterpreted in the Western Go World?

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